SME
Kinematics of Mechanisms · Problem type
Time ratio of a quick-return linkage
Use this for “A crank-rocker’s crank turns 204° on the working stroke and 156° on the return (β = 24°) at 60 rpm. Find the time ratio and each stroke’s time.”
Chart
0.5667 + 0.4333 = 1 s.
Offset angle β = 24°
Time ratio Q = 1.3077
Crank speed n = 60 rpm
This is an example. Type your own values to start a new problem.
Q
=
180°
+
°
180°
−
°
n
Crank speed
example · rpm
T
Time per turn
worked out (not typed)
tw
Working-stroke time
calculated
tᵣ
Return time
calculated
Formulas
Q
=
180 +
β
180 −
β
1.3077
=
180 + 24
180 − 24
T
=
60
n
1
=
60
60
tw
=
(
180 +
β
360
)
×
T
0.5667
=
(
180 + 24
360
)
×
1
tᵣ
=
(
180 −
β
360
)
×
T
0.4333
=
(
180 − 24
360
)
×
1
Assumptions
•
The crank turns at a steady speed, so time on each stroke follows the angle it sweeps.
•
β is the angle between the crank’s two positions at the rocker’s ends, less 180°.
•
β = 0 for a centred linkage: both strokes take the same time.
Step-by-step
WE KNOW
β = 24°, n = 60 rpm
FIND
time ratio (Q), time per turn (T), working-stroke time (tw), return time (tᵣ)
Step 1 · Find time ratio (Q)
Use Q = (180 + β) ÷ (180 − β)
The
working
stroke
sweeps
180°
+
β
and
the
return
180°
−
β
,
at
the
same
crank
speed.
Q
=
180 + 24
180 − 24
Q
=
204
156
Q
=
1.3077
Step 2 · Find time per turn (T)
Use T = 60 ÷ n
One turn takes 60 seconds over the turns per minute.
T
=
60
60
T
=
1
s
Step 3 · Find working-stroke time (tw)
Use tw = ((180 + β) ÷ 360) × T
The working stroke’s share of a turn, times the time per turn.
tw
=
(
180 + 24
360
)
×
1
tw
=
204
360
tw = 0.5667 s
Step 4 · Find return time (tᵣ)
Use tᵣ = ((180 − β) ÷ 360) × T
The return stroke’s share of a turn, times the time per turn.
tᵣ
=
(
180 − 24
360
)
×
1
tᵣ
=
156
360
tᵣ = 0.4333 s
Check
Put the numbers back in. Both sides should match.
1.3077
=
180 + 24
180 − 24
✓
1
=
60
60
✓
0.5667
=
(
180 + 24
360
)
×
1
✓
0.4333
=
(
180 − 24
360
)
×
1
✓
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